91Ó°ÊÓ

There are $18$red numbers and $18$odd numbers.

The sample space of the $38$possible outcomes in the given game is,

Let it be s.

$S=\left\{\begin{array}{l}0,00,1,2,3,4,5,6,7,8\\ 9,10,11,12,13,14,15\\ 16,17,18,19,20,21,22\\ 23,24,25,26,27,28,29\\ 30,31,32,33,34,35,36\end{array}\right\}$

There are $18$red numbers and $18$odd numbers.

The probability of $P\left(\text{red}\right)$is

$$\begin{gathered} P\left(red\right)=\frac{n\left(red\right)}{n\left(s\right)} \\ n\left(\text{red}\right)=18;n\left(S\right)=38 \\ P\left(red\right)=\frac{18}{38} \\ P\left(red\right)=0.47 \end{gathered}$$

There are $18$red numbers and $18$odd numbers.

The probability of $P\left(1^{st}12\right)$is

$$\begin{gathered} P\left(1^{st}12\right)=\frac{n\left(1^{st}12\right)}{n\left(S\right)} \\ n\left(1^{st}12\right)=12;n\left(S\right)=38 \\ P\left(1^{st}12\right)=\frac{12}{38} \\ P\left(1^{st}12\right)=0.32 \end{gathered}$$

There are $18$ red numbers and $18$ odd numbers.

The probability of $P\left(evennumber\right)$is

$$\begin{gathered} P\left(\text{even number}\right)=\frac{n\left(\text{evennumber}\right)}{n\left(S\right)} \\ n\left(\text{evennumber}\right)=18;n\left(S\right)=38 \\ P\left(\text{even number}\right)=\frac{18}{38} \\ P\left(\text{even number}\right)=0.47 \end{gathered}$$

There are $18$ red numbers and $18$ odd numbers.

Getting an odd number is not the complement of getting an even number because we have $38$outcomes in the sample space which includes $0$and $00$.

There are $18$ red numbers and $18$ odd numbers.

To get the mutually exclusive events, there should not be any common terms in between the two events.

Therefore, the two mutually exclusive events are odd and even numbers or in other words, it is red and black numbers.

There are $18$ red numbers and $18$ odd numbers.

To get two independent even, we have to show the following.

$$\begin{gathered} P(\text{Even}âˆ\pounds \text{1stdozen})=P\left(\text{Even}\right) \\ P(\text{lstdozen}âˆ\pounds \text{Even})=P\left(1\text{stdozen}\right) \\ P\left(\text{Even AND 1stdozen}\right)=P\left(\text{Even}\right)P\left(1\text{stdozen}\right) \end{gathered}$$

Therefore,

$$\begin{gathered} P(\text{Even}âˆ\pounds 1\text{stdozen})=0.5;P\left(\text{Even}\right)=0.47 \\ P(\text{Even}âˆ\pounds 1\text{stdozen})≠P\left(\text{Even}\right) \end{gathered}$$